The $(x,y)$ coordinates of the stationary point of a function of 2 variables $f(x,y)$ are given by solving
the following 2 equations for $x$ and $y$
\\begin{align}
\\partial f \\over \\partial x &= 0 \\\\[1em]
\\partial f \\over \\partial y &= 0
\\end{align}
In this case you get two equations to solve for $x$ and $y$
\n\\begin{align}
\\simplify[std]{{-2*b}*(x-{c})*e^(-(x-{c})^2-(y-{d})^2)} &= 0 \\\\[1em]
\\simplify[std]{{-2*b}*(y-{d})*e^(-(x-{c})^2-(y-{d})^2)} &= 0
\\end{align}
We can cancel off the term $\\simplify[std]{e^(-(x-{c})^2-(y-{d})^2)}$ in both equations as $\\simplify[std]{e^(-(x-{c})^2-(y-{d})^2)} \\neq 0,\\;\\forall x,\\;y$.
\nOn solving these, we get
\n\\[ x = \\var{c}, \\quad \\;y=\\var{d} \\]
\nSo the stationary point is $(\\var{c},\\var{d}) \\in D$.
\nOn substituting these values into $f(x,y)$ we get:
\n\\[ f(\\var{c},\\var{d})=\\simplify[std,!zeropower,!othernumbers]{{a}+{b}*e^0={a+b}} \\]
", "preamble": {"js": "", "css": ""}, "name": "Luis's copy of Find the stationary point of a function on a disk", "variable_groups": [], "extensions": [], "functions": {}, "variablesTest": {"maxRuns": 100, "condition": ""}, "statement": "In the following question find the $(x,y)$ coordinates of the single stationary point $(a,b) \\in D$ of the continuous function $f: D \\rightarrow \\mathbb{R}$:
\n\\[f(x,y) = \\simplify[std]{{a} + {b}*e^(-(x-{c})^2-(y-{d})^2)}\\]
\nwhere
\n\\[D = \\{(x,y): \\simplify[std]{(x-{c})^2+(y-{d})^2}\\} \\le \\var{r}\\]
\nthat is, $D$ is a disk of radius $\\simplify[std]{sqrt({r})}$ and centre $(\\var{c},\\var{d})$.
\nInput both cooordinates as fractions or integers and not decimals.
", "parts": [{"variableReplacements": [], "stepsPenalty": 0, "showFeedbackIcon": true, "prompt": "$x$ – coordinate, $a=$ [[0]]
\n$y$ – coordinate, $b=$ [[1]]
\nInput value of $f(x,y)$ at $(a,b)$:
\n$f(a,b)=$ [[2]]
\nIf you want some help, click on Show steps.
", "steps": [{"variableReplacements": [], "showFeedbackIcon": true, "prompt": "\n \n \nThe $(x,y)$ coordinates of the stationary point of a function of 2 variables $f(x,y)$ are given by solving
the following 2 equations for $x$ and $y$
\\[\\begin{eqnarray*}\n \n \\partial f \\over \\partial x &=&0\\\\\n \n \\\\\n \n \\partial f \\over \\partial y &=&0\n \n \\end{eqnarray*}\n \n \\]
\n \n \n \nIn this case you get two equations to solve for $x$ and $y$
\n \n \n ", "variableReplacementStrategy": "originalfirst", "scripts": {}, "type": "information", "showCorrectAnswer": true, "marks": 0}], "variableReplacementStrategy": "originalfirst", "type": "gapfill", "scripts": {}, "gaps": [{"mustBeReduced": false, "variableReplacements": [], "showFeedbackIcon": true, "scripts": {}, "correctAnswerStyle": "plain", "type": "numberentry", "marks": 2, "showCorrectAnswer": true, "allowFractions": false, "correctAnswerFraction": false, "minValue": "{c}", "variableReplacementStrategy": "originalfirst", "notationStyles": ["plain", "en", "si-en"], "maxValue": "{c}", "mustBeReducedPC": 0}, {"mustBeReduced": false, "variableReplacements": [], "showFeedbackIcon": true, "scripts": {}, "correctAnswerStyle": "plain", "type": "numberentry", "marks": 2, "showCorrectAnswer": true, "allowFractions": false, "correctAnswerFraction": false, "minValue": "{d}", "variableReplacementStrategy": "originalfirst", "notationStyles": ["plain", "en", "si-en"], "maxValue": "{d}", "mustBeReducedPC": 0}, {"mustBeReduced": false, "variableReplacements": [], "showFeedbackIcon": true, "scripts": {}, "correctAnswerStyle": "plain", "type": "numberentry", "marks": 1, "showCorrectAnswer": true, "allowFractions": false, "correctAnswerFraction": false, "minValue": "{a+b}", "variableReplacementStrategy": "originalfirst", "notationStyles": ["plain", "en", "si-en"], "maxValue": "{a+b}", "mustBeReducedPC": 0}], "showCorrectAnswer": true, "marks": 0}], "ungrouped_variables": ["a", "c", "b", "d", "s3", "s2", "s1", "s5", "s4", "r"], "metadata": {"description": "Find the coordinates of the stationary point for $f: D \\rightarrow \\mathbb{R}$: $f(x,y) = a + be^{-(x-c)^2-(y-d)^2}$, $D$ is a disk centre $(c,d)$.
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